In a Young's double slit experiment, `I_0` is the intensity at the central maximum and `beta` is the fringe width. The intensity at a point P distant x from the centre will be

Options

A. (a) `I_ocos(pix)/(beta)`
B. (b) `4I_ocos^2(pix)/(beta)`
C. (c) `I_ocos^2(pix)/(beta)`
D. (d) `I_o/4cos^2(pix)/(beta)`

Answer

C

Explanation

Path difference at point `P=(xd)/(D)`
Phase difference at point `P=(2pi)/(lambda)(xd)/(D)=(2pix)/(beta)`
`I_0=4I_1`, intensity at point P
`I=I_1+I_1+2I_1co s(2pix)/(beta)=2I_1[1+cos(2pix)/(beta)]`
`=I_0cos^2(pix)/(beta)`